A particle moves along the $x$ -axis. The function $v(t)$ gives the particle's velocity at any time $t>0$ : $v(t)=t^4-8t^2+20$ What is the particle's velocity $v(t)$ at $t=2$ ? $v(2)=$
Answer: In the first part, we need to find the particle's velocity. We have a function for the particle's velocity, so all we have to do is evaluate the function. In the second part, we need to find the particle's acceleration. Since acceleration is the rate of change of velocity, we need to find the derivative of $v(t)$. In other words, if $a(t)$ gives the particle's acceleration at any time $t>0$, then $a(t)=v'(t)$. First, let's evaluate $v(2)$ : $\begin{aligned} v({2})&=({2})^4-8({2})^2+20 \\\\ &=4 \end{aligned}$ Notice that the velocity is positive. This means the particle is moving to the right. Now, let's differentiate $v(t)$ to find $a(t)$ : $\begin{aligned} a(t)&=v'(t) \\\\ &=\dfrac{d}{dt}[t^4-8t^2+20] \\\\ &=4t^3-16t \end{aligned}$ To find the particle's acceleration at $t=2$, we need to evaluate $a(2)$. $\begin{aligned} a({2})&=4({2})^3-16({2}) \\\\ &=0 \end{aligned}$ Since the acceleration is zero, we know the particle is neither speeding up nor slowing down. The particle's velocity at $t=2$ is $4$. The particle's acceleration at $t=2$ is $0$. At $t=2$, the particle is neither speeding up nor slowing down.